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$$\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$

How can I prove this statement?

Michael Hardy
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Kuba
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    Sometimes you can just google for your anwser.. http://www.cut-the-knot.org/triangle/SinCosFormula.shtml – Nils Ziehn Jan 11 '14 at 21:05
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    Have you tried to do it or look it up? –  Jan 11 '14 at 21:05
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    It depends on what definition you want to start with. It's pretty straightforward if you can use Euler's formula. But you're probably looking for an elementary geometric proof. – Ayman Hourieh Jan 11 '14 at 21:06
  • Here's a great thread on this topic. http://math.stackexchange.com/questions/1292/how-can-i-understand-and-prove-the-sum-and-difference-formulas-in-trigonometry – littleO Jan 11 '14 at 23:53

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$e^{i(a+b)}=\cos(a+b)+i\sin(a+b)$ by Euler's formula. But $e^{i(a+b)}=e^{ia}e^{ib}=(\cos(a)+i\sin(a))(\cos(b)+i\sin(b))= \cos(a)\cos(b)-\sin(a)\sin(b)+i(\sin(a)\cos(b) + \cos(a)\sin(b))$

So by comparing real and imaginary parts you obtain the trigonometric addition formulae for both $\sin$ and $\cos$.

Edward ffitch
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    This seems like a very strange choice of proof. It relies on the validity of Euler's formula. IMHO, it simply leaves the reader needing to prove Euler's formula. There are some very simple proofs which rely on first principles. – Fly by Night Jan 11 '14 at 21:18
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    @FlybyNight If you define $\sin$, $\cos$, and the exponential by their power series, then Euler's theorem is simple to prove and this is actually the simplest proof. You have to examine what your first principles are. – Ryan Reich Jan 11 '14 at 22:04
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    @RyanReich Come on Ryan, this is just silly. The trigonometric functions are rooted in geometry. I realise that power series definitions are useful and worthwhile, but they're formal shadows of the true spirit of trigonometry. Using infinite series and complex numbers to prove something that can be proved using circles and triangles is totally unnecessary. – Fly by Night Jan 11 '14 at 22:09
  • Except if you use the taylor definition, you need to prove derivative of sin(x) is cos(x). Is there another way other than applying the limit definition to prove the derivative (which requires use of sin(a+b) formula)? – 1110101001 Jan 11 '14 at 22:13
  • @user2612743 It's sort of trivial using power series. – Ryan Reich Jan 11 '14 at 22:14
  • But to find the taylor series you need the derivative of a function, right? And unless there is another way, to get the derivative of sin(x) requires use of the sum identity with the limit definition of the derivative. – 1110101001 Jan 11 '14 at 22:16
  • @user2612743 My point is that you take the infinite series to be the definition of these functions. This is an approach used to provide a rigorous foundation for these geometric arguments. – Ryan Reich Jan 11 '14 at 22:17
  • @FlybyNight I would argue that since complex numbers are essentially geometric in character, this very proof is the one that connects the "formal shadow" with the "true spirit" of trigonometry. It is both rigorous and intuitive. – Ryan Reich Jan 11 '14 at 22:18
  • @RyanReich I would agree that the complex numbers are a beautiful intersection of geometry and algebra. My problem is that the proof requires far more mathematical machinery than necessary. I think proofs should be as simple as possible. The more simple, the more beautiful. One would need to write a small book to take a high school student to the Euler formula proof. I'd need a stick and some wet sand. – Fly by Night Jan 11 '14 at 22:26
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    @FlybyNight If you're saying it depends on your audience, I agree. A naive audience prefers a naive proof; a sophisticated audience prefers a more rigorous proof even if it means sacrificing the original intuition. A really sophisticated audience then loops back around and considers the naive proof acceptable because they know how to make it rigorous. And then come the debates over whether intuition is foundational or emergent. There's no one way to do math, is what I'm saying, even if you know everything. – Ryan Reich Jan 11 '14 at 22:47
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    @RyanReich I would agree with almost everything you say except that increased complexity does not necessarily give increased rigour. The proof that $\tfrac{\mathrm{d}}{\mathrm{d}\theta}\sin\theta = \cos\theta$ depends on a simple, geometrical argument. If simple proof mean less rigour then calculus is in trouble. Anyway. I don't want to seem argumentative. I really do understand your point of view. Whenever possible, I try to show my students how new results "boil down" to well-know results as a special case. – Fly by Night Jan 11 '14 at 23:00
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I typed "Proof of Trigonometric formulae" in to Google and the second hit was an extensive Wikipedia article which supplies proofs to many, many trigonometric identities.

Click here for the section that you want.

Fly by Night
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  • You're right, I'm not using english in Google mostly, and in my language I couldn't find it. Thanks for your help :) – Kuba Jan 11 '14 at 21:14
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    Not to get off topic, but I often find explanations on math.stackexchange to be more enlightening than what I find on wikipedia or elsewhere. For this particular question, very good threads already exist on math.stackexchange. – littleO Jan 11 '14 at 21:19
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    @littleO One may vote to close a question as a duplicate. Have a look at the bottom of the questions and you will see the words share-edit-close-flag. Click Close and choose the appropriate category. – Fly by Night Jan 11 '14 at 21:21
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This is the way I learned it: a geometric proof like this on Wikipedia.

The segment $OP$ has length $1$. We have then, $\sin(\alpha + \beta) = PB = PR + RB = \cos(\alpha) \sin(\beta) + \sin(\alpha) \cos(\beta)$.

dwarandae
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This gem comes from E. Schmidt: Consider $f(x)=\sin(\alpha+\beta-x)\cos(x)+\cos(\alpha+\beta-x)\sin(x)$. Since $f'(x)=0$ we know that $f$ is constant, hence $f(0)=f(\beta)$.

Michael Hoppe
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Here is a modern proof using sleight of hand.(1)An isometry of the Euclidean plane $E$ is a function $f:E\to E$ that preserves the distance $d$ between points: $d(A,B)=d(f(A),f(B))$ for all $A,B\in E$. We have (2):An isometry $f$ is onto.Because, for $Q\in E$, let $A_1,A_2,A_3\in E$ be non-co-linear with $Q\not \in \{f(A_1),f(A_2),f(A_3)\}$.The unique $P$ such that $d(P,A_i)=d(Q,f(A_i))$ for $i=1,2,3$ must satisfy $f(P)=Q$.We have (3):An isometry that fixes 3 non-co-linear points $A_1,A_2,A_3$ is the identity function.Because,as in (2),for any $P$, the only $Q$ that can satisfy $ d(P,A_i)=d(Q,f(A_i))=d(Q,A_i)$ for $i=1,2,3$ is $Q=P$.We have (4):If isometries $ f,g$ agree on 3 non-co-linear points then $f=g$. Because $g^{-1}:E\to E$ exists by (2), and hence is also an isometry (obviously), and by (3) we have $g^{-1}(f(P))=P$ for all $ P$, so $$g(P)=g(g^{-1}(f(P))=f(P).$$ Now to use all this :(5): Choose a pair of orthogonal co-ordinate axes for $E$ with origin $O$. Consider two types of isometry :First, $R_b$ is a counter-clockwise rotation about $O$ through angle $b$. Second, $M_b$ sends the point with co-ordinates $(x,y)$ to $$(x\cos b-y\sin b,x\sin b+y\cos b).$$The fact that $M_b$ is an isometry relies only on the theorem of Pythagoras, or in other words, $\cos^2 b+\sin^2 b=1$. Now for any $b$, observe that $R_b$ and $M_b$ agree on the 3 non-co-linear points $(0,0),(1,0),(0,1)$.Therefore by (4) we have $$R_b=M_b\text { for all b }.$$Finally, since $R_b$ and $R_c$ are rotations about $O$ we have$$R_bR_c=R_{b+c} \text{ for all b,c }.$$ Therefore we have $$M_{b+c}=R_{b+c}=R_bR_c=M_bM_c\text{ for all b,c}.$$ By comparing the co-ordinates of $M_{b+c}(1,0)$ with the co-ordinates of $M_bM_c(1,0)$, we have the angle-sum formulas for $\cos$ and $\sin$.

DanielWainfleet
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